1. Copyright © Cengage Learning. All rights reserved. CHAPTER 8 RELATIONS

2. Copyright © Cengage Learning. All rights reserved. Equivalence Relations SECTION 8.3

3. 3 The Relation Induced by a Partition

4. 4 A partition of a set A is a finite or infinite collection of nonempty, mutually disjoint subsets whose union is A. The diagram of Figure 8.3.1 illustrates a partition of a set A by subsets A 1, A 2,..., A 6. Figure 8.3.1 A Partition of a Set

5. 5 The Relation Induced by a Partition

6. 6 Example 1 – Relation Induced by a Partition Let A = {0, 1, 2, 3, 4} and consider the following partition of A: {0, 3, 4}, {1}, {2}. Find the relation R induced by this partition. Solution: Since {0, 3, 4} is a subset of the partition, 0 R 3 because both 0 and 3 are in {0, 3, 4}, 3 R 0 because both 3 and 0 are in {0, 3, 4},

7. 7 Example 1 – Solution 0 R 4 because both 0 and 4 are in {0, 3, 4}, 4 R 0 because both 4 and 0 are in {0, 3, 4}, 3 R 4 because both 3 and 4 are in {0, 3, 4}, and 4 R 3 because both 4 and 3 are in {0, 3, 4}. Also, 0 R 0 because both 0 and 0 are in {0, 3, 4} 3 R 3 because both 3 and 3 are in {0, 3, 4}, and 4 R 4 because both 4 and 4 are in {0, 3, 4}. cont’d

8. 8 Example 1 – Solution Since {1} is a subset of the partition, 1 R 1 because both 1 and 1 are in {1}, and since {2} is a subset of the partition, 2 R 2 because both 2 and 2 are in {2}. Hence R = {(0, 0), (0, 3), (0, 4), (1, 1), (2, 2), (3, 0), (3, 3), (3, 4), (4, 0), (4, 3), (4, 4)}. cont’d

9. 9 The Relation Induced by a Partition The fact is that a relation induced by a partition of a set satisfies all three properties: reflexivity, symmetry, and transitivity.

10. 10 Definition of an Equivalence Relation

11. 11 Definition of an Equivalence Relation A relation on a set that satisfies the three properties of reflexivity, symmetry, and transitivity is called an equivalence relation. Thus, according to Theorem 8.3.1, the relation induced by a partition is an equivalence relation.

12. 12 Example 2 – An Equivalence Relation on a Set of Subsets Let X be the set of all nonempty subsets of {1, 2, 3}. Then X = {{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}} Define a relation R on X as follows: For all A and B in X, A R B ⇔ the least element of A equals the least element of B. Prove that R is an equivalence relation on X.

13. 13 Example 2 – Solution R is reflexive: Suppose A is a nonempty subset of {1, 2, 3}. [We must show that A R A.] It is true to say that the least element of A equals the least element of A. Thus, by definition of R, A R A. R is symmetric: Suppose A and B are nonempty subsets of {1, 2, 3} and A R B. [We must show that B R A.] Since A R B, the least element of A equals the least element of B. But this implies that the least element of B equals the least element of A, and so, by definition of R, B R A.

14. 14 Example 2 – Solution R is transitive: Suppose A, B, and C are nonempty subsets of {1, 2, 3}, A R B, and B R C. [We must show that A R C.] Since A R B, the least element of A equals the least element of B and since B R C, the least element of B equals the least element of C. Thus the least element of A equals the least element of C, and so, by definition of R, A R C. cont’d

15. 15 Equivalence Classes of an Equivalence Relation

16. 16 Equivalence Classes of an Equivalence Relation Suppose there is an equivalence relation on a certain set. If a is any particular element of the set, then one can ask, “What is the subset of all elements that are related to a?” This subset is called the equivalence class of a.

17. 17 Equivalence Classes of an Equivalence Relation When several equivalence relations on a set are under discussion, the notation [a] R is often used to denote the equivalence class of a under R. The procedural version of this definition is

18. 18 Example 5 – Equivalence Classes of a Relation Given as a set of Ordered Pairs Let A = {0, 1, 2, 3, 4} and define a relation R on A as follows: R = {(0, 0), (0, 4), (1, 1), (1, 3), (2, 2), (3, 1), (3, 3), (4, 0), (4, 4)}. The directed graph for R is as shown below. As can be seen by inspection, R is an equivalence relation on A. Find the distinct equivalence classes of R.

19. 19 Example 5 – Solution First find the equivalence class of every element of A. Note that [0] = [4] and [1] = [3]. Thus the distinct equivalence classes of the relation are {0, 4}, {1, 3}, and {2}.

20. 20 Equivalence Classes of an Equivalence Relation The first lemma says that if two elements of A are related by an equivalence relation R, then their equivalence classes are the same. This lemma says that if a certain condition is satisfied, then [a] = [b]. Now [a] and [b] are sets, and two sets are equal if, and only if, each is a subset of the other.

21. 21 Equivalence Classes of an Equivalence Relation Hence the proof of the lemma consists of two parts: first, a proof that [a]  [b] and second, a proof that [b]  [a]. To show each subset relation, it is necessary to show that every element in the left-hand set is an element of the right-hand set. The second lemma says that any two equivalence classes of an equivalence relation are either mutually disjoint or identical.

22. 22 Equivalence Classes of an Equivalence Relation The statement of Lemma 8.3.3 has the form if p then (q or r), where p is the statement “A is a set, R is an equivalence relation on A, and a and b are elements of A,” q is the statement “[a]  [b] = Ø,” and r is the statement “[a] = [b].”

23. 23 Congruence Modulo n

24. 24 Example 10 – Equivalence Classes of Congruence Modulo 3 Let R be the relation of congruence modulo 3 on the set Z of all integers. That is, for all integers m and n, Describe the distinct equivalence classes of R. Solution: For each integer a,

25. 25 Example 10 – Solution Therefore, In particular, cont’d

26. 26 Example 10 – Solution Now since 3 R 0, then by Lemma 8.3.2, More generally, by the same reasoning, Similarly, cont’d

27. 27 Example 10 – Solution And Notice that every integer is in class [0], [1], or [2]. Hence the distinct equivalence classes are cont’d

28. 28 Example 10 – Solution In words, the three classes of congruence modulo 3 are (1) the set of all integers that are divisible by 3, (2) the set of all integers that leave a remainder of 1 when divided by 3, and (3) the set of all integers that leave a remainder of 2 when divided by 3. cont’d

29. 29 Congruence Modulo n

30. 30 Example 11 – Evaluating Congruences Determine which of the following congruences are true and which are false. a. b. c. Solution: a. True. 12 – 7 = 5 = 5  1. Hence 5 | (12 – 7), and so 12  7 (mod 5). b. False. 6 – (–8) = 14, and because 14  4  k for any integer k. Consequently, c. True. 3 – 3 = 0 = 7  0. Hence 7 | (3 – 3), and so 3  3 (mod 7).

31. 31 A Definition for Rational Numbers

32. 32 A Definition for Rational Numbers For a moment, forget what you know about fractional arithmetic and look at the numbers as symbols. Considered as symbolic expressions, these appear quite different. In fact, if they were written as ordered pairs (1, 3) and (2, 6) they would be different.

33. 33 A Definition for Rational Numbers The fact that we regard them as “the same” is a specific instance of our general agreement to regard any two numbers as equal provided the cross products are equal: ad = bc. This can be formalized as follows, using the language of equivalence relations.

34. 34 Example 12 – Rational Numbers Are Really Equivalence Classes Let A be the set of all ordered pairs of integers for which the second element of the pair is nonzero. Symbolically, Define a relation R on A as follows: For all (a, b), (c, d)  A, The fact is that R is an equivalence relation. a. Prove that R is transitive. b. Describe the distinct equivalence classes of R.

35. 35 Example 12(a) – Solution Suppose (a, b), (c, d), and (e, f ) are particular but arbitrarily chosen elements of A such that (a, b) R (c, d ) and (c, d ) R (e, f ). [We must show that for all (a, b), (c, d ), (e, f )  A, if (a, b) R (c, d ) and (c, d ) R (e, f ), then (a, b) R (e, f ).] [We must show that (a, b) R (e, f ).] By definition of R, Since the second elements of all ordered pairs in A are nonzero, b  0, d  0, and f  0.

36. 36 Example 12(a) – Solution Multiply both sides of equation (1) by f and both sides of equation (2) by b to obtain Thus and, since d  0, it follows from the cancellation law for multiplication that It follows, by definition of R, that (a, b) R (e, f ) [as was to be shown]. cont’d

37. 37 Example 12(b) – Solution There is one equivalence class for each distinct rational number. Each equivalence class consists of all ordered pairs (a, b) that, if written as fractions a/b, would equal each other. The reason for this is that the condition for two rational numbers to be equal is the same as the condition for two ordered pairs to be related. cont’d

38. 38 Example 12(b) – Solution For instance, the class of (1, 2) is since and so forth. cont’d